MATHEMATICS: PAPER II MARKING GUIDELINES

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NATIONAL SENIOR CERTIFICATE EXAMINATION NOVEMBER 017 MATHEMATICS: PAPER II MARKING GUIDELINES Time: hours 150 marks These marking guidelines are prepared for use by examiners and sub-examiners, all

1 NATIONAL SENIOR CERTIFICATE EXAMINATION NOVEMBER 017 MATHEMATICS: PAPER II MARKING GUIDELINES Time: hours 150 marks These marking guidelines are prepared for use by examiners and sub-examiners, all of whom are required to attend a standardisation meeting to ensure that the guidelines are consistently interpreted and applied in the marking of candidates' scripts. The IEB will not enter into any discussions or correspondence about any marking guidelines. It is acknowledged that there may be different views about some matters of emphasis or detail in the guidelines. It is also recognised that, without the benefit of attendance at a standardisation meeting, there may be different interpretations of the application of the marking guidelines. IEB Copyright 017

2 NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II MARKING GUIDELINES Page of 11 SECTION A QUESTION 1 (a) 0,77 (The second mark awarded for rounding off correctly) () (b) C (1) (c) A = 0,668 B = 0,064 () (d) No, you would be extrapolating (The second mark is for the concept of extrapolation) () [7] QUESTION (a) 4 0 m OA = = 0 tan AOB ˆ = AOB ˆ = 6,4 (4) (b) m = 1 Midpoint of OA = (1;) y = 1 x + c Subs (1; ) = 1 + c y = 1 x + 5 (4) (c) x = (1) (d) y = 1 () + 5 y = 1 (x ) + (y 1) = r Subs (0;0) (0 ) + (0 1) = r r = 10 (x ) + (y 1) = 10 (5) [15] IEB Copyright 017

3 NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II MARKING GUIDELINES Page of 11 QUESTION (a) (1) sin(1 + ) = k = sin 5 = k () cos( ) = sin 5 = k (1) () () cos(75 ) = cos 5 = 1 k (There is a method mark for workings.) () (b) cos θ cos θ sin θ sin θ cos θ sin θ 1 cos θ sin sin θ sin θ θ 1 cos θ+sin θ sin θ sin θ +cos θ cos θ +sin θ sin θ sin θ sin θ sin θ therefore LHS = RHS (6) (c) sin θ sin θ = 0 sin θ ( sin θ ) = 0 sin θ = 0 θ = 0 + k 180 Alternate: θ = 0 + k 60 OR θ = k 60 OR sin θ = θ= 41,8 + k 60 OR θ =18, + k 60 K Z (6) [18] IEB Copyright 017

4 NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II MARKING GUIDELINES Page 4 of 11 QUESTION 4 (a) M(; 1) (1) (b) (0 ) + (y + 1) = 5 y + y 15 = 0 (y + 5)(y ) = 0 y = 5 OR y = C(0;) () (c) m CM 1 4 = = 0 m AC = 4 y = x + () 4 (d) 0 = + 4 x x = 4 A( 4; 0) (x ) + (0 + 1) = 5 (x ) = 4 x = ± 4 AB = 4 units 1,9 units AB =,1 units (4) [11] IEB Copyright 017

5 NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II MARKING GUIDELINES Page 5 of 11 QUESTION 5 (a) R.T.P: ˆ ˆ CAE = ABC Construction: refer to diagram for the construction Proof: OAC ˆ + CAE ˆ = 90 (Tangent perpendicular to line through centre) FCA ˆ = 90 (Angles in semi-circle) OFC ˆ + OAC ˆ = 90 (Angles in triangle) therefore OFC ˆ = CAE ˆ but OFC ˆ = ABC ˆ (Angles in same segment) therefore CAE ˆ = ABC ˆ (7) (b) B ˆ = 70 F ˆ = 5 G ˆ ˆ 1+ G = 70 (Angles in same segment) (Exterior angle of cyclic quad equal to the interior opposite angle) F ˆ ˆ = G = 5 (tan chord theorem) G ˆ 1 = 18 (5) [1] D B O F A C E IEB Copyright 017

6 NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II MARKING GUIDELINES Page 6 of 11 QUESTION 6 (a) A = 50 (1) (b) 400 (1) (c) P = 50 and M = 100 () (d) Q = greater than 00 and less than 5 (approximate) Q 1 = 00 IQR = 110 (ca mark based on values) () (e) Mean = 50 () (f) (1) It would stay the same. It is only the upper 5% of data that are affected. () Or It would decrease. People leave the contract therefore less people. () Standard deviation would decrease. The difference between the new mean and the data would decrease. () () It would skew the data to the left. The vales above the median are less spread out. or mean < median () [15] 77 marks IEB Copyright 017

7 NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II MARKING GUIDELINES Page 7 of 11 SECTION B QUESTION 7 (a) m OA = Equation of line OA is y = x Equation of line EF is y + x = 10 (x) + x = 10 7x = x = (This represents the height of the triangle.) 7 y = 0 7 Coordinates of point E y + 0 = 10 E(0; 5) y = Area of EBO = Area of EBO = units (8) 7 (b) C(4; 0) (0) + x = 10 x = Area of DCF = Area of DCF = units (4) 5 [1] IEB Copyright 017

8 NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II MARKING GUIDELINES Page 8 of 11 QUESTION 8 (a) (1) OC () () B (6; ) () () m OC = = 1 OCB ˆ = 45 COB ˆ = 90 angles of a Alternate solution: coscob COB ˆ 90 CAB ˆ 45 (Angle at centre) ˆ CAB ˆ = 45 (angle at centre) (4) (b) Circumference = r Circumference = 18 units or 6,66 units COB ˆ = 60 (Angle at centre = x angle at circumference) The size of angle θ after B moves into new position 9 1 = sin θ θ = 0 θ = = 150 (Area rule) B needs to move 90 anti-clockwise Therefore Point B needs to move OR Point B needs to move 6, 66 units. (6) [14] IEB Copyright 017

9 NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II MARKING GUIDELINES Page 9 of 11 QUESTION 9 (a) (b) Ĉ is a common angle D ˆ = A ˆ (tan chord theorem) Therefore ΔADC ΔDBC (A.A.A) OR B ˆ ˆ = ADC (Angles in a triangle) (4) DC AC = BC DC DC = AC.BC but AC = AB + BC Therefore DC = BC (AB + BC) DC = AB.BC + BC ( ΔADC ΔDBC) AB.BC = DC BC (4) [8] QUESTION 10 (a) ADL ˆ = 90 ACB ˆ = 90 Therefore (Angle in semi-circle) (one mark for the reason) DL CB (Converse: corresponding angles are equal) (4) (b) LC = LA (radii of the large circle) SD = SL = SA but LA = SA + SL therefore (radii of small circle) Alternative solution: AD = DC (converse: midpoint DL BC) In ΔACL DS CL (midpoint theorem) LC = SD LC = SD () (c) AS = SL and AL = LB; radii SL = 1 AB 4 () (d) LB = 15 units (radius) 9 LM = (prop theorem) LM = 8,44 units () [1] IEB Copyright 017

10 NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II MARKING GUIDELINES Page 10 of 11 QUESTION 11 (a) (1) BE = OA ED (radii) () () AE = EC (Line from centre is perpendicular to chord) BE = BC EC (OA ED) = BC AE (4) (b) Construction DB DBE ˆ =BDE ˆ (Tangents drawn from common chord) ˆ ˆ 180 θ DBE = BDE = ˆ 180 θ A = (tan chord theorem) ˆ C θ = (Opp angles of cyclic quad) ˆ 180 θ θ C = = 90 + A B C D E (6) [1] IEB Copyright 017

11 NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II MARKING GUIDELINES Page 11 of 11 QUESTION 1 (a) Front view of the circles 6 h h = 6 h = Height of B (Pythagoras) + 6 (5) (b) (1) sin 50 = 11, AB 11, AB = sin 50 AB = 14,6 metres (4) () A view of the triangle made by the two pieces of rope and the horizontal plane sin A sin 70 = 1 14,6 A ˆ = 56,68 B ˆ = 5, OPTION 1 EA 14,6 = sin 5, sin 70 OPTION EA = 1,48 metres EA = ,6 (1)(14,6)cos5, EA = 1,48 metres (6) [15] 7 marks Total: 150 marks IEB Copyright 017

MATHEMATICS: PAPER II MARKING GUIDELINES

NATIONAL SENIOR CERTIFICATE EXAMINATION NOVEMBER 05 MATHEMATICS: PAPER II MARKING GUIDELINES Time: 3 hours 50 marks These marking guidelines are prepared for use by examiners and sub-examiners, all of